Cho hàm số \(y = f\left( x \right) = \left\{ \begin{array}{l}3{x^2}\;{\rm{khi}}\;0 \le x \le 1\\4 - x\;{\rm{khi}}\;1 < x \le 2\end{array} \right.\). Tính tích phân \(\int\limits_0^2 {f\left( x \right)dx} \).

Đáp án đúng là: A

\(\int\limits_0^2 {f\left( x \right)dx} = \int\limits_0^1 {f\left( x \right)dx} + \int\limits_1^2 {f\left( x \right)dx} \)\( = \int\limits_0^1 {3{x^2}dx} + \int\limits_1^2 {\left( {4 - x} \right)dx} \)\( = \left. {{x^3}} \right|_0^1 + \left. {\left( {4x - \frac{{{x^2}}}{2}} \right)} \right|_1^2\)\( = 1 + 6 - \frac{7}{2} = \frac{7}{2}\).